# Momentum Conservation

1. Jan 4, 2006

### metalmagik

Hello, I do not know how to do this momentum conservation problem in the vertical direction, when the object is to find how far the second object went.
A gun is fired vertically into a 1.40-kg block of wood at rest directly above it. If the bullet has a mass of 21.0g (.021 kg) and a speed of 210 m/s, how high will the block rise into the air after the bullet becomes embedded in it?
I have found momentum which was 44.1 kgm/s and the velocity of the block which was 3.15 m/s but I just do not know how I would find the distance the block traveled. The answer is .491 m. Quick help is appreciated, thank you.

2. Jan 4, 2006

### Tide

HINT: After the collision, mechanical energy is conserved.

3. Jan 4, 2006

### metalmagik

I'm still a little confused, I've tried doing mgh = 1/2mv^2 + mgh but I realized I do not have the starting height...I am lost to what else I am supposed to do regarding this problem.

4. Jan 4, 2006

### dicerandom

Just define your coordinate system so that the starting height of the block (i.e. the height of the table, or whatever it's on) is zero. Then when you solve for h that will be the height it reached off the table.

5. Jan 4, 2006

### metalmagik

Thank you, I have solved this problem, it was really just a simple interpretation problem on my part, thanks again, I have one more question I do not quite understand that I expect to see on my exam tomorrow, I will post it in a new thread called "Momentum Particles" if anyone is kind enough to help me with it before the end of tonight.